Question

Using differentials, find the approximate value of the following up to 3 places of decimal.
$(26{)}^{\frac{1}{3}}$

Answer 1

$\left(26{)}^{\frac{1}{3}}Letf\right(x)=x^{\frac{1}{3}}\Rightarrow f^{\prime }(x)=\frac{1}{3}{x}^{\frac{-2}{3}}$
Let $x=27and\Delta x=-1,$
Now, $f(x+\Delta x)\simeq f\left(x\right)+\Delta f^{\prime }\left(x\right)\Rightarrow (x+\Delta x{)}^{\frac{1}{3}}\simeq x^{\frac{1}{3}}+\frac{\Delta x}{3x^{\frac{2}{3}}}\Rightarrow (27-1{)}^{\frac{1}{3}}\simeq (27{)}^{\frac{1}{3}}+\frac{(-1)}{3(27{)}^{\frac{2}{3}}}=3-\frac{1}{3\left\{\right(27{)}^{\frac{1}{3}}{\}}^2}=3-\frac{1}{3\times 3^2}=3-0.037=2.963\Rightarrow (26{)}^{\frac{1}{3}}\simeq \mathrm{2.963.}$